What EMF is induced in solenoid with L=6π×10⁻⁵H when current changes by 2A in 3.14s?

ε=L·dI/dt=6π×10⁻⁵×2/π=12×10⁻⁵V. So α=12.

What is the self-inductance of this solenoid?

L=μ₀n²lA=4π×10⁻⁷×10⁶×0.3×5×10⁻⁴=6π×10⁻⁵H.

Solenoid 30cm, 10 turns/cm, A=5cm², I changes 2A→4A in 3.14s — EMF induced

JEERANKERS

PhysicsEM Induction

QA 30 cm long solenoid has 10 turns per cm and area of 5 cm². The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is $\alpha\times10^{-5}$ V. The value of $\alpha$ is:

A) 60 B) 12 C) 120 D) 34Electromagnetic Induction

MCQ

✅ Correct Answer

B) α=12

✓

Solution

Find self-inductance L

$n=10\text{ turns/cm}=1000\text{ turns/m}$
$l=0.30\text{ m}$, $A=5\times10^{-4}\text{ m}^2$

$L=\mu_0 n^2 lA=4\pi\times10^{-7}\times(10^3)^2\times0.3\times5\times10^{-4}$
$=4\pi\times10^{-7}\times10^6\times1.5\times10^{-4}$
$=4\pi\times10^{-7}\times1.5\times10^2$
$=6\pi\times10^{-5}\text{ H}$

Compute induced EMF

$\varepsilon=L\dfrac{dI}{dt}=6\pi\times10^{-5}\times\dfrac{4-2}{3.14}$

$=6\pi\times10^{-5}\times\dfrac{2}{\pi}$ (using $3.14\approx\pi$)

$=6\times10^{-5}\times2=12\times10^{-5}\text{ V}$

$\therefore\quad\alpha=\boxed{12}$

📘 Self-Inductance of Solenoid

Self-inductance $L=\mu_0 n^2 lA$ where $n$ is turns per unit length, $l$ is length, $A$ is cross-section area. Induced EMF $=L\cdot dI/dt$. Note: when $3.14s$ appears and $L$ contains $\pi$, they cancel cleanly.

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Important Concepts

Self-Inductance FormulaL=μ₀n²lA. n=1000 turns/m, l=0.3m, A=5×10⁻⁴m². L=4π×10⁻⁷×10⁶×0.3×5×10⁻⁴=6π×10⁻⁵H.

Faraday's Law EMFε=L·dI/dt. dI=4−2=2A, dt=3.14s=π s. ε=6π×10⁻⁵×2/π=12×10⁻⁵V.

Why 3.14s?The problem uses t=3.14s≈π specifically so that π cancels in the calculation, giving a clean integer answer α=12.

Solenoid vs Single LoopFor a solenoid with N total turns: L=μ₀N²A/l=μ₀n²lA. Total turns N=n×l=1000×0.3=300 turns.

FAQs

What is n in turns/m?

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n=10 turns/cm = 10×100 turns/m = 1000 turns/m.

How to compute L step by step?

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L=4π×10⁻⁷ × (1000)² × 0.3 × 5×10⁻⁴ = 4π×10⁻⁷ × 10⁶ × 1.5×10⁻⁴ = 4π×10⁻⁷ × 150 = 600π×10⁻⁷ = 6π×10⁻⁵ H.

What is the back-EMF physically?

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When current increases, the solenoid's magnetic flux increases, inducing an EMF that opposes the change (Lenz's law). This back-EMF = L·dI/dt.

What is the total flux linkage?

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Nφ=L×I=6π×10⁻⁵×3=18π×10⁻⁵ Wb (at I=3A average).

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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